Tìm x thuộc Z:
\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\)
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\(\dfrac{2x-9}{240}\) \(=\) \(\dfrac{39}{80}\) \(\Leftrightarrow\) \(\left(2x-9\right)\) \(.80\) \(=\) \(240.39\)
\(2x\) \(-\) \(9\) \(=\) ( \(240\) \(.\) \(39\) ) \(:\) \(80\)
\(2x\) \(-\) \(9\) \(=\) \(117\)
\(2x\) \(=\) \(126\)
\(x\) \(=\) \(63\)
\(\dfrac{2x-9}{240}=\dfrac{117}{240}\)
=> 2x -9=117
2x=117+9
2x=126
x=126/2
x=63
\(\dfrac{2x-9}{240}\)=\(\dfrac{117}{240}\)
►2x-9=117
2x=117+9
2x=126
x=126:2
x=63
Vậy:x=63
\(\frac{2x-9}{240}=\frac{39}{80}\) với x \(\in\) Z
Vì : \(240:80=3\) . Vậy : \(\Rightarrow2x-9\) gấp 3 lần 39 .
\(\Rightarrow\) Tổng của \(2x-9=39.3\)
\(2x-9=117\)
\(2x=117+9\)
\(2x=126\)
\(x=126:2\)
\(x=63\)
a: \(B=\dfrac{3x\left(2x-3\right)-4\left(2x+3\right)-4x^2+23x+12}{\left(2x-3\right)\left(2x+3\right)}\cdot\dfrac{2x+3}{x+3}\)
\(=\dfrac{6x^2-9x-8x-12-4x^2+23x+12}{2x-3}\cdot\dfrac{1}{x+3}\)
\(=\dfrac{2x^2+6x}{\left(2x-3\right)}\cdot\dfrac{1}{x+3}=\dfrac{2x}{2x-3}\)
b: 2x^2+7x+3=0
=>(2x+3)(x+2)=0
=>x=-3/2(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{2\cdot\left(-2\right)}{-2-3}=\dfrac{-4}{-7}=\dfrac{4}{7}\)
d: |B|<1
=>B>-1 và B<1
=>B+1>0 và B-1<0
=>\(\left\{{}\begin{matrix}\dfrac{2x+2x-3}{2x-3}>0\\\dfrac{2x-2x+3}{2x-3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3< 0\\\dfrac{4x-3}{2x-3}>0\end{matrix}\right.\Leftrightarrow x< \dfrac{3}{4}\)
\(\frac{2x-9}{240}=\frac{39}{80}\Leftrightarrow\left(2x-9\right).80=240.39\)
\(2x-9=\left(240.39\right):80\)
\(2x-9=117\)
\(2x=126\)
\(x=63\)
2x-9 / 240 = 39 / 80
2x-9 / 240 = 117 / 240
2x - 9 = 117
2x = 117 + 9
2x = 126
x = 126 : 2
x = 63
Vậy x = 63
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Rightarrow\frac{2x-9}{240}=\frac{117}{240}\)
\(\Rightarrow2x-9=117\)
\(\Rightarrow2x=117+9\)
\(\Rightarrow2x=126\)
\(\Rightarrow x=126:2\)
\(\Rightarrow x=63\)
Vậy \(x=63\)
Chúc bạn học tốt !!!
Ta có: \(\frac{2x-9}{240}=\frac{39}{80}\)
=> \(\frac{2x-9}{240}=\frac{117}{240}\)
=> 2x-9 = 117
=> 2x = 117+9
=> 2x = 126
=> x = 126:2
=> x = 63
\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\)\(\Rightarrow\)\(2x-9\)=\(\dfrac{240.39}{80}\)=\(117\)
\(2x-9=117\)\(\Rightarrow\)\(2x=117+9=126\)\(\Rightarrow\)\(x=126:2=63\)
\(\dfrac{2x-9}{240}\)=\(\dfrac{39}{80}\) \(\Rightarrow\) \(2x-9=\)\(\dfrac{240.39}{80}\)=117
\(2x-9=117\)\(\Rightarrow\)\(2x=117-9=108\)\(\Rightarrow\)\(x=108:2=54\)